Your "order 2" remark I guess generalises to something like: if $R$ is a commutative ring with $2$ not a zero divisor, and if $x^2=1$ only has a finite number of solutions in $R$ (e.g. if $R$ is an integral domain then both these conditions are satisfied), then $SL(3,R)$ has more elements of order 2 then $SL(2,R)$ so there's no injection. [Hint: use Cayley-Hamilton on a matrix of order 2 to convince yourself it must be scalar, if I got it right; conversely use the fact that $-1$ isn't 1 to build more elements in $SL(3,R)$ than this]
–
Kevin BuzzardNov 26 '10 at 9:44

3

Alex: If you have any doubts about Cayley-Hamilton for a 2x2 matrix over an arbitrary commutative ring, why not just bash out the algebra!
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Kevin BuzzardNov 26 '10 at 10:01

1

Alex: for the general case, consider the fact that you believe it for the polynomial ring $R=Z[A_{11},A_{12},...,A_{nn}]$ (which is an ID) and now apply it to the matrix whose $(i,j)$th entry is $A_{ij}$ and see what this tells you.
–
Kevin BuzzardNov 26 '10 at 10:08

4

Actually, these comments make a good answer to another question currently kicking around MO: "what is the point of a universal object?". An $n\times n$ matrix over a general ring $S$ is a map from that ring $R$ above to $S$, reducing questions like CH for any ring, to CH for that one universal ring. One notices this when doing the first exercise I suggested: checking CH in the 2x2 case by bashing out the algebra. One has a bunch of equations in $A,B,C,D$ all of which magically work out, but instead of thinking of $A,B,C,D$ as varying through the ring one can think of them as poly variables.
–
Kevin BuzzardNov 26 '10 at 10:27

1

@BS: One might also need something like: "the map $SL(3,R)\to SL(2,R)$ extends to a map of rings $M(3,R)\to M(2,R)$" for this strategy to work, right? That doesn't sound so good (e.g. I don't think it's true for maps $SL(3,R)\to SL(3,R)$---things like "inverse-transpose" give trouble...)...
–
Kevin BuzzardNov 26 '10 at 12:04

The elements of $R$ which square to zero form a nilpotent ideal $I$. It follows that if $G$ embeds into $\text{SL}_2(R)$, the image of any elements of order $4$ in $\text{SL}_2(R/I)$ must, by the above computation, have order dividing $2$. In particular the image of $G$ is not isomorphic to $G$, so it must be trivial since $G$ is simple. Hence the image of $G$ in $\text{SL}_2(R)$ consists only of matrices congruent to the identity $\bmod I$. But any such matrix has trace squaring to zero, hence order dividing $4$, which contradicts the existence of elements of order $7$ in $G$. So no such embedding exists.

Case:$2$ is nilpotent in $R$. It is still true in this case that $SL_3({\mathbb F}_2)$ embeds into $\text{SL}_3(R)$, because the two complex 3-dimensional irreducible representations of $SL_3({\mathbb F}_2)=\text{PSL}_2({\mathbb F}_7)$ are realisable over ${\mathbb Z}_2$ (they need $(1\pm\sqrt{-7})/2$ which are in ${\mathbb Z}_2$), and $R$ is a ${\mathbb Z}_2$-algebra. However, as explained in the follow-up question, $SL_3({\mathbb F}_2)$ is not a subgroup of $\text{SL}_2(R)$ for any $R$.

Case:$2$ is not a zero divisor in $R$. As has been observed elsewhere, in this case $G = S_4$ embeds into $\text{SL}_3(R)$. Let $g \in \text{SL}_2(R)$ be an element of order dividing $2$ and trace $r$. Then $g^2 = 1$ and $g^2 - rg + 1 = 0$, hence $rg = 2$. Squaring gives $r^2 = 4$, so $r$ is also not a zero divisor. It follows that $g$ must be a scalar multiple of the identity, hence central. But $S_4$ contains elements of order $2$ which are not central. (As does $S_3$. Hence this argument also shows, as in Tim Dokchitser's answer, that $S_3$ does not embed into $\text{SL}_2(R)$ in this case.)

In particular the above arguments cover the case that $2$ is invertible, as well as the case that $R$ is an integral domain. The remaining case is that $2 \neq 0$, and it is a zero divisor but is not nilpotent.

One may hope that the reduction to Artinian rings as in the follow-up question and the fact that we know the answer both when $2$ is invertible and for ${\mathbb Z}_2$-algebras can actually finish this off.

@Qiaochu: Neat! I changed "order 4" to "order dividing 4", hope it's ok (to make sure that "necessary and sufficient" is clear). Feel free to roll back if you don't like this.
–
Tim DokchitserNov 28 '10 at 16:27

1

Perhaps this is useful: the complex 3-dimensional representation of $SL_3({\mathbb Z}/2)=PSL_2(7)$ is realisable over ${\mathbb Z}_2$ (it needs $(1+\sqrt{-7})/2$ which is in ${\mathbb Z}_2$). So this group embeds into $SL_3(R)$ whenever $2$ is nilpotent in $R$, not just when $2=0$.
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Tim DokchitserNov 28 '10 at 17:35

Here is a brute force proof that $SL_3(R)$ does not embed into $SL_2(R)$ for any commutative ring $R$ with 1 where 2 is invertible, inspired by Kevin's universality remarks and Alex's observation that $S_3$ in my comment does not land in $SL_2(R)$.

The claim is that the symmetric group $S_3$ cannot be embedded in $SL_2(R)$ for any $R$: suppose it can be, take 2 general matrices $M=\begin{pmatrix}a&b\cr c&d\end{pmatrix}$ and $T=\begin{pmatrix}e&f\cr g&h\end{pmatrix}$ and consider the relations $M^3=id=T^2$, $M^2T=TM$ and $det M=det T=1$. These are 4+4+4+1+1=14 polynomial relations in 8 variables $a,...,h$, and a Groebner basis computation shows that the ideal they generate is
$\langle h^2-1,g,f,h-e,d-1,c,b,a-1\rangle$; in Mathematica this is

In other words, the relations imply that $a=d=1$ and $b=c=0$ for any $R$, so $M=1$, contradicting the assumption that $S_3\to SL(2,R)$ is injective.

On the other hand, using
$$
M=\begin{pmatrix}0&0&1\cr 1&0&0\cr 0&1&0\end{pmatrix},\qquad
T=\begin{pmatrix}0&-1&0\cr -1&0&0\cr 0&0&-1\end{pmatrix},
$$
we can embed $S_3$ into $SL(3,R)$ for any ring $R$.

P.S. Hopefully, there is a better proof that $S_3$ does not embed into $SL(2,R)$!

Edit: The Groebner basis computation works over $Z[1/2]$, so this only works for rings $R$ with 2 invertible. Kevin & John: thank you for pointing this out!

Tim: I've not had time to think about it properly, but here's an idea which might show $S_3$ can't map into $SL(2,R)$: (1) check true for $R$ a field (char not 2,3 is fine; char 2 and char 3 are special cases but this theory is well-understood---just not by me). (2) Deduce for integral domains. (3) By considering maps induced by $R\to R/P$ deduce some statement of the form "map to SL(2,R) must degenerate mod P for all P and hence (because intersection of all primes is nilradical) must land in nilradical in some way and then (4) reduce to case where nilradical is itself nilpotent and then hope!
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Kevin BuzzardNov 26 '10 at 22:00

Stop! In trying to push through the above strategy I notice that $S_3$ does embed into $SL(2,Z/2Z)$ so something is wrong. Did you do the Groebner basis calc over Q??
–
Kevin BuzzardNov 26 '10 at 22:03

1

Either I am not thinking right at all today or something is wrong here - isn't $SL(2,Z/2Z)$ isomorphic with $S_3$?
–
John ShareshianNov 26 '10 at 22:08

@John: right! My (admittedly limited) experience with Groebner bases calculations on computers is that they are typically only implemented over fields, which makes me think that Tim's calculation was done over a field, probably the rationals. So probably he has proved that there is no $\mathbf{Q}$-algebra $R$ with the property we seek.
–
Kevin BuzzardNov 26 '10 at 22:39

@Kevin&John: Oops, you're absolutely right, it must be that Mathematica does it over $Q$. I'll try to check that. To be continued...
–
Tim DokchitserNov 26 '10 at 22:49

Define a sequence of groups $G_i$ and associated group rings $R_i=\mathbb{Q}[G_i]$. To start put $G_0=\mathbb{Q}$. Then define $G_{i+1}=SL_3(R_i)$. The group $SL_3(R_i)$ is a subgroup of $SL_2(R_{i+1})$ because $G_{i+1}$ is a subgroup of $SL_2(R_{i+1})$ (as the group of certain diagonal matrices). Similarly, $G_i$ is a subgroup of $G_{i+1}$, and hence $R_i$ is a subring of $R_{i+1}$. Then $R=\bigcup_i R_i$ is the ring you want.

I think the OP assumes $R$ to be commutative, but I might be wrong. Any way what do you mean by $SL_n$ of a non-commutative ring ? The only thing I can think of is the subgroup $E_n(R)$ of $GL_n$ generated by "elementary matrices".
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BS.Nov 26 '10 at 11:29

@BS: You are right. There is no good definition of SL_2 for noncommutative rings, and my answer is rubbish.
–
Boris BukhNov 26 '10 at 11:44

1

Presumably there is a good notion of $GL_2$ though? Something like: $R$-linear automorphisms of $R^2$? So perhaps you can give a $GL_2$ example?
–
Kevin BuzzardNov 26 '10 at 11:49

1

No Kevin $GL_2(R)\cong GL_3 (R)$ for a noncomm ring without IBN. It is a no brainer...
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Bugs BunnyNov 26 '10 at 12:34

1

Yes, IBN may fail: take a ring $R$ generated by coefficients $x_{ij}$ and $y_{ji}$ of generic $mxn$ and $nxm$ matrices $X$ and $Y$ with relations $XY=1$ and $YX=1$. All you need to show is that $R$ is not zero.
–
Bugs BunnyNov 29 '10 at 15:46

My guess that it should not be possible because $SL_2(R)$ would not have a 3-soluble nonnilpotent subgroup. I am not sure whether it is true that any soluble non-nilpotent subgroup would lie in Borel subgroup but I imagine that this is right...

I agree that it should be possible to turn this into a proof. However, even when $R$ is an algebraically closed field, a soluble group is contained in a Borel only up to finite index (e.g. consider $S=N_G(T)$, where $G$ has a solvable Weyl group).
–
GuntramNov 26 '10 at 13:06

1

I don't know what 3-soluble means either :-/ but if $R$ is the $p$-adic integers then the congruence subgroup $\Gamma(p)$---doesn't it have the property that if you keep taking commutator subgroups then you get smaller and smaller without ever reaching 1? So I am a bit concerned about the claims in the answer if $R=Z/p^nZ$ for some big $n$...
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Kevin BuzzardNov 26 '10 at 13:41

More stupid comment: $SL(2,R)$ is soluble if $R$ is quite small (e.g. Artin local with residue field of order 2 should do it, right?)
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Kevin BuzzardNov 26 '10 at 14:05

I see where things have gone wrong but they fixable: all your examples are nilpotent... I still believe that $SL_2 (R)$ would contain no 3-soluble non-nilpotent subgroup...
–
Bugs BunnyNov 29 '10 at 15:49

EDIT: I misquoted the book, and the answers here show it isn't fixable.

Let $A,B \in SL_2(R)$. This book by Brumfiel and Hilden has the following two facts in it:

$Tr(A)Tr(B) = Tr(AB) + Tr(AB^{-1})$.

The pair $A,B$ is uniquely determined, up to simultaneous conjugation, by the elements $Tr(A)$, $Tr(B)$, and $Tr(AB)$.

Taking $A^2 = 1$ and $A = B$, the first fact implies that $Tr(A) = \pm 2$, and then the second fact implies that $A$ is conjugate (and hence equal) to $\pm Id$. Since there is an embedding $S_4 \hookrightarrow SL_3(R)$ (which is described in the other answers), we can finish by noting that the images of the transpositions are not central in $SL_3(R)$.

I would guess that there's a more elementary way to prove that involutions in $SL_2(R)$ are central, but I don't know one at the moment.

Assume that $\SL_3(R)$ is a subgroup of $\SL_2(R)$. We wish to obtain a contradiction.

Here is the strategy. Suppose that $R$ contains a subring of the form $A \oplus B$
where $2A = 0$. Then $SL_3(\F_2)$ is a subgroup of $\SL_3(A)$, which is a subgroup
of $\SL_3(A \oplus B)$, which is a subgroup of $\SL_3(R)$. Hence, under our assumption on $R$, $\SL_3(\F_2)$ is a subgroup of $\SL_2(R)$, and this is ruled out by Silence Dogood's answer. $R$ trivially admits such
a decomposition when $2 = 0$. Hence we may assume that $2 \ne 0$, and thus that
$S_4$ is a subgroup of $\SL_3(R)$, and hence of $\SL_2(R)$.

If $S \subset R$ contains a subring of the form $A \oplus B$ with $2A = 0$, then
so does $R$.
Thus, WLOG, assume that $R$ is generated by the entries of $g-1$ where $g \in S_4
\subset \SL_2(R)$.
Let $K \subset S_4$ denote the Klein $4$-subgroup. Then any map $S_4 \rightarrow G$ is injective if and only if the restriction $K \rightarrow G$ is non-zero (Obvious). $K$ is the only non-trivial normal subgroup of $S_4$ which is a $p$-group (Obvious). By construction, $R$ is Noetherian. If $x \in R$ is any element, and $\m$ is a maximal ideal containing the annihilator of $x$, then $x$ is non-zero in the localization $R_{\m}$. Hence there exists an $\m$ such that $K \rightarrow \SL_2(R_{\m})$ is non-zero, so $S_4
\rightarrow \SL_2(R_{\m})$ is injective. (Choose $x$ to be a non-zero matrix entry of $g-1$ for $g \in K$.) Let $A = R_{\m}$, and let $k = A/\m$. Consider the projection map $S_4 \rightarrow \SL_2(k)$, and let $H$ denote the kernel. Let $g$ be an element of $H$ which is not the identity (if such an element exists). By the Krull intersection theorem (as in SD's answer), there exists a minimal integer $n$ such that $$g - 1 \equiv 0 \mod \m^n, \qquad (g - 1) \not\equiv 0 \mod \m^{n+1}.$$ If $i$ is co-prime to the characteristic of $k$, then it is a unit in $A$, and $$g^i - 1 = (1 + (g-1))^i - 1 \equiv i (g-1) \mod \m^{n+1} \not\equiv 0 \mod \m^{n+1}.$$ It follows that the order of $g$ is some power of the characteristic (or is trivial if $\mathrm{char}(k) = 0$), and hence $H$ is a $p$-group. Hence either $S_4$ injects into $\SL_2(k)$, or $k$ has characteristic $2$ and $H = K$. The former does not occur. We shall prove that $2 = 0$ in $A$. The image of $S_4$ in $\SL_2(k)$ is $S_3$. $S_4$ contains an element $M$ of order $2$ which maps to an element of order $2$ in $S_3$ (for example, any $2$-cycle). The matrix $M$ has order two, and hence satisfies the polynomial $M^2 - 1 = 0$. Yet $M$ also has determinant one, and thus also satisfies the polynomial $M^2 - \mathrm{trace}(M) M + 1 = 0$, by Cayley--Hamilton. It follows that $\mathrm{trace}(M) M = 2 \ne 0$ (by assumption). Yet $M$ has at least one entry that is a unit, and thus $(\mathrm{trace}(M)) = (2)$ in $A$, and it follows easily that the image of $M$ is a scalar matrix in $\SL_2(k)$. Since $k$ has characteristic $2$, this implies that the image of $M$ in $\SL_2(k)$ is trivial ($v^2 = 1$ implies that $v = 1$), a contradiction. Hence $2 = 0$ in $A$.

We have now shown that $2 = 0$ in $A = R_{\m}$. Suppose we can show in
addition that $A$ has finite length, that is $A/\m^k = A$ for some $k$. Assume this is so.
Let $x_1, \ldots, x_n$ be generators of $\m^k \subset R$. By definition, $x_i$ maps to zero
in the localization map $R \rightarrow R_{\m} = A$. Thus there exists an element
$y_i \notin \m$ such that $y_i x_i = 0$. Let $y = y_1 \times \ldots
\times y_n$. Since $y_i \notin \m$, the product $y \notin \m$. It follows that
$$y + \m^k = R,$$
as the ideal on the LHS is not contained in any maximal ideal. On the other hand,
$y$ annihilates $\m^k$ by construction. Thus, by the Chinese remainder theorem,
$$R = R/y \m^k = R/y \oplus R/\m^k = R/y \oplus A.$$

Since $2 = 0$ in $A$, this shows that $R$ has the required decomposition.
Thus we will be done if we can show that $A$ has finite length.
Equivalently, we are done if we can show that the non-unit elements of $A$
are nilpotent.

@Jack: I think your group lands in $GL$ rather than $SL$, the generators have determinant $-1=3$.
–
Tim DokchitserDec 4 '10 at 19:06

1

@frogger: my girlfriend is on call this weekend and I've got all the kids. I noticed the post but didn't yet read it. I'm really pleased the question is answered but didn't just want to mindlessly upvote until I'd read what you had done. [which I will, soon...]
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Kevin BuzzardDec 5 '10 at 18:37

This argument does point out some strange (to me) things in the Artinian local case: Let G(n) be SL(2,ZZ[x]/(n,xx)). I would have expected G(4) to be more flexible in all ways than G(2), but in fact they each appear to have strengths. G(2) contains dihedral groups of order 6, 8, and 12, and the symmetric group S4, while G(4) contains none of those. On the other hand, G(4) contains elements of 8 and elementary abelian groups of rank 6, while G(2) only has elements of order 4 and subgroups of rank 3. Is there some sense in which G(4) is more abelian?
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Jack SchmidtDec 5 '10 at 20:03

1

@Jack: good question, but one that, slightly more out of desire to keep these comments from growing too convoluted and get it better attention than to create as many spin-offs from my first question as possible, I think you should ask as its own question :)
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Jonathan KiehlmannDec 5 '10 at 21:04

Assume that $2\neq 0$ in $R$ - this is the only assumption I will need. All this is based on ideas of Kevin Buzzard, Tim Dokchitser and Kevin Ventullo.

Since $1\neq-1\in R$, there is an embedding of $S_4$ into $SL_3(R)$:
$$
(1,2,3,4)\mapsto \begin{pmatrix}0&1&0\\\\-1&0&0\\\\0&0&1 \end{pmatrix},\;\;
(1,2)\mapsto \begin{pmatrix}-1&0&0\\\\0&0&1\\\\0&1&0 \end{pmatrix}.
$$

As Kevin notes, a matrix of order 2 in $SL_2(R)$ is either a scalar (minimal polynomial linear) or the minimal and therefore also the characteristic polynomial is $x^2-1$. Since the involutions in $S_4$ are not central, the latter must be the case for all of them (we are assuming that $R$ is commutative, otherwise we don't know how to make sense of $SL$, so all scalar matrices are central). But the characteristic polynomial of a matrix in $SL_2$ must have constant term $1\neq-1$ - contradiction.

If $2=0$ in $R$, then we can embed $SL_3(\mathbb{F}_2)\cong PSL_2(\mathbb{F}_7)$ in $SL_3(R)$. This group is simple, so no element embeds as scalars into $SL_2(R)$. Thus, the minimal polynomials of all the elements of $SL_3(\mathbb{F}_2)\subseteq SL_2(R)$ are quadratic. Can anyone see how to finish this?

This does look promising! And if 1=-1 then can one perhaps use SL(3,Z/2)?
–
Kevin BuzzardNov 26 '10 at 17:32

1

Suppose -1 does not have a square root in R. Then I think any element g of order 4 in SL_2(R) must have characteristic polynomial g^2 = -1, but S_4 contains elements of order 4 whose squares are different.
–
Qiaochu YuanNov 26 '10 at 17:49

I hate to be the guy that only points out errors, but... shouldn't the characteristic polynomial of $M$ be $(x-a)(x-d)-bc$?
–
Kevin VentulloNov 27 '10 at 22:22

Ah, but it's OK! If the char. poly has constant term $-1$, then the det. is $-1$, so we must have $-1=1$.
–
Kevin VentulloNov 27 '10 at 22:41

1

@Alex: an explicit example of something I'm worried about (i.e. worried you're saying I'm asserting) is that if $R=Z/4Z$ and $g$ is the matrix $[1,2;0,1]$then $g$ is not scalar and the char poly of $g$ is not $x^2-1$ either.
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Kevin BuzzardNov 28 '10 at 15:09